CHEMISTRY
The Central Science
9th Edition
Chapter 17
Additional Aspects of Aqueous
Equilibria
David P. White
Prentice Hall © 2003
Chapter 17
The Common Ion Effect
• The solubility of a partially soluble salt is decreased
when a common ion is added.
• Consider the equilibrium established when acetic acid,
HC2H3O2, is added to water.
• At equilibrium H+ and C2H3O2- are constantly moving
into and out of solution, but the concentrations of ions is
constant and equal.
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Chapter 17
The Common Ion Effect
• Consider the addition of C2H3O2-, which is a common
ion. (The source of acetate could be a strong electrolyte
such as NaC2H3O2.)
• Therefore, [C2H3O2-] increases and the system is no
longer at equilibrium.
• So, [H+] must decrease.
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Chapter 17
Buffered Solutions
Composition and Action of Buffered
Solutions
• A buffer consists of a mixture of a weak acid (HX) and
its conjugate base (X-):
HX(aq)
H+(aq) + X-(aq)
• The Ka expression is
[H  ][ X - ]
Ka 
[HX]
[HX]

[H ]  K a
[X ]
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Chapter 17
Buffered Solutions
Composition and Action of Buffered
Solutions
• A buffer resists a change in pH when a small amount of
OH- or H+ is added.
• When OH- is added to the buffer, the OH- reacts with HX
to produce X- and water. But, the [HX]/[X-] ratio
remains more or less constant, so the pH is not
significantly changed.
• When H+ is added to the buffer, X- is consumed to
produce HX. Once again, the [HX]/[X-] ratio is more or
less constant, so the pH does not change significantly.
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Chapter 17
Buffered Solutions
•
•
•
•
•
Buffer Capacity and pH
Buffer capacity is the amount of acid or base neutralized
by the buffer before there is a significant change in pH.
Buffer capacity depends on the composition of the buffer.
The greater the amounts of conjugate acid-base pair, the
greater the buffer capacity.
The pH of the buffer depends on Ka.
If Ka is small (i.e., if the equilibrium concentration of
undissociated acid is close to the initial concentration),
then
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Chapter 17
Buffered Solutions
Buffer Capacity and pH
[HX]

 log[ H ]   log K a  log
[X ]
[X- ]
 pH  pK a  log
[HX]
Addition of Strong Acids or Bases to Buffers
• We break the calculation into two parts: stoichiometric
and equilibrium.
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Chapter 17
Buffered Solutions
Addition of Strong Acids or Bases to
Buffers
• The amount of strong acid or base added results in a
neutralization reaction:
X- + H3O+  HX + H2O
HX + OH-  X- + H2O.
• By knowing how must H3O+ or OH- was added
(stoichiometry) we know how much HX or X- is formed.
• With the concentrations of HX and X- (note the change in
volume of solution) we can calculate the pH from the
Henderson-Hasselbalch equation.
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Chapter 17
Buffered Solutions
Addition of Strong Acids or Bases to
Buffers
conjugate base
pH  pK a  log
acid
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Chapter 17
Acid-Base Titrations
Strong Acid-Strong Base Titrations
• A plot of pH versus volume of acid (or base) added is
called a titration curve.
• Consider adding a strong base (e.g. NaOH) to a solution
of a strong acid (e.g. HCl).
– Before any base is added, the pH is given by the strong acid
solution. Therefore, pH < 7.
– When base is added, before the equivalence point, the pH is
given by the amount of strong acid in excess. Therefore, pH <
7.
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Chapter 17
Acid-Base Titrations
Strong Acid-Strong Base Titrations
− At equivalence point, the amount of base added is
stoichiometrically equivalent to the amount of acid originally
present. Therefore, the pH is determined by the salt solution.
Therefore, pH = 7.
• Consider adding a strong base (e.g. NaOH) to a solution
of a strong acid (e.g. HCl).
• We know the pH at equivalent point is 7.00.
• To detect the equivalent point, we use an indicator that
changes color somewhere near 7.00.
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Chapter 17
Acid-Base Titrations
Strong Base-Strong Acid
Titrations
Acid-Base Titrations
•
•
•
•
•
Strong Acid-Strong Base Titrations
The equivalence point in a titration is the point at which
the acid and base are present in stoichiometric quantities.
The end point in a titration is the observed point.
The difference between equivalence point and end point
is called the titration error.
The shape of a strong base-strong acid titration curve is
very similar to a strong acid-strong base titration curve.
Initially, the strong base is in excess, so the pH > 7.
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Chapter 17
Acid-Base Titrations
Strong Acid-Strong Base Titrations
• As acid is added, the pH decreases but is still greater than
7.
• At equivalence point, the pH is given by the salt solution
(i.e. pH = 7).
• After equivalence point, the pH is given by the strong
acid in excess, so pH < 7.
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Chapter 17
Acid-Base Titrations
Weak Acid-Strong Base Titrations
• Consider the titration of acetic acid, HC2H3O2 and
NaOH.
• Before any base is added, the solution contains only weak
acid. Therefore, pH is given by the equilibrium
calculation.
• As strong base is added, the strong base consumes a
stoichiometric quantity of weak acid:
HC2H3O2(aq) + NaOH(aq)  C2H3O2-(aq) + H2O(l)
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Chapter 17
Acid-Base Titrations
Weak Acid-Strong Base Titrations
• There is an excess of acid before the equivalence point.
• Therefore, we have a mixture of weak acid and its
conjugate base.
– The pH is given by the buffer calculation.
• First the amount of C2H3O2- generated is calculated, as well as the
amount of HC2H3O2 consumed. (Stoichiometry.)
• Then the pH is calculated using equilibrium conditions.
(Henderson-Hasselbalch.)
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Chapter 17
Prentice Hall © 2003
Chapter 17
Acid-Base Titrations
Weak Acid-Strong Base Titrations
• At the equivalence point, all the acetic acid has been
consumed and all the NaOH has been consumed.
However, C2H3O2- has been generated.
– Therefore, the pH is given by the C2H3O2- solution.
– This means pH > 7.
• More importantly, pH  7 for a weak acid-strong base titration.
• After the equivalence point, the pH is given by the strong
base in excess.
• The equivalence point is determined by Ka of the acid.
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Chapter 17
Prentice Hall © 2003
Chapter 17
Acid-Base Titrations
•
•
•
•
Weak Acid-Strong Base Titrations
For a strong acid-strong base titration, the pH begins at
less than 7 and gradually increases as base is added.
Near the equivalence point, the pH increases
dramatically.
For a weak acid-strong base titration, the initial pH rise is
more steep than the strong acid-strong base case.
However, then there is a leveling off due to buffer effects.
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Chapter 17
Acid-Base Titrations
Weak Acid-Strong Base Titrations
• The inflection point is not as steep for a weak acid-strong
base titration.
• The shape of the two curves after equivalence point is the
same because pH is determined by the strong base in
excess.
• Two features of titration curves are affected by the
strength of the acid:
– the amount of the initial rise in pH, and
– the length of the inflection point at equivalence.
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Chapter 17
Acid-Base Titrations
Titrations of Polyprotic Acids
• In polyprotic acids, each ionizable proton dissociates in
steps.
• Therefore, in a titration there are n equivalence points
corresponding to each ionizable proton.
• In the titration of H3PO3 with NaOH.
– The first proton dissociates to form H2PO3-.
– Then the second proton dissociates to form HPO32-.
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Chapter 17
Solubility Equilibria
The Solubility-Product Constant, Ksp
• Consider
BaSO4(s)
• for which
Ba2+(aq) + SO42-(aq)
K sp  [Ba 2 ][SO24- ]
• Ksp is the solubility product. (BaSO4 is ignored because
it is a pure solid so its concentration is constant.)
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Chapter 17
Solubility Equilibria
The Solubility-Product Constant, Ksp
• In general: the solubility product is the molar
concentration of ions raised to their stoichiometric
powers.
• Solubility is the amount (grams) of substance that
dissolves to form a saturated solution.
• Molar solubility is the number of moles of solute
dissolving to form a liter of saturated solution.
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Chapter 17
Solubility Equilibria
Solubility and Ksp
• To convert solubility to Ksp
• solubility needs to be converted into molar solubility (via
molar mass);
• molar solubility is converted into the molar concentration
of ions at equilibrium (equilibrium calculation),
• Ksp is the product of equilibrium concentration of ions.
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Chapter 17
Factors that Affect
Solubility
The Common Ion Effect
• Solubility is decreased when a common ion is added.
• This is an application of Le Châtelier’s principle:
CaF2(s)
Ca2+(aq) + 2F-(aq)
• as F- (from NaF, say) is added, the equilibrium shifts
away from the increase.
• Therefore, CaF2(s) is formed and precipitation occurs.
• As NaF is added to the system, the solubility of CaF2
decreases.
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Chapter 17
Factors that Affect
Solubility
Solubility and pH
• Again we apply Le Châtelier’s principle:
CaF2(s)
Ca2+(aq) + 2F-(aq)
– If the F- is removed, then the equilibrium shifts towards the
decrease and CaF2 dissolves.
– F- can be removed by adding a strong acid:
F-(aq) + H+(aq)
HF(aq)
– As pH decreases, [H+] increases and solubility increases.
• The effect of pH on solubility is dramatic.
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Chapter 17
Factors that Affect
Solubility
Formation of Complex Ions
• A Consider the formation of Ag(NH3)2+:
Ag+(aq) + 2NH3(aq)
Ag(NH3)2(aq)
• The Ag(NH3)2+ is called a complex ion.
• NH3 (the attached Lewis base) is called a ligand.
• The equilibrium constant for the reaction is called the
formation constant, Kf:
[Ag(NH 3 )2 ]
Kf 
[Ag  ][ NH3 ]2
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Chapter 17
Factors that Affect
Solubility
Formation of Complex Ions
• Consider the addition of ammonia to AgCl (white
precipitate):
AgCl(s)
Ag+(aq) + Cl-(aq)
Ag+(aq) + 2NH3(aq)
Ag(NH3)2(aq)
• The overall reaction is
AgCl(s) + 2NH3(aq)
Ag(NH3)2(aq) + Cl-(aq)
• Effectively, the Ag+(aq) has been removed from solution.
• By Le Châtelier’s principle, the forward reaction (the
dissolving of AgCl) is favored.
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Chapter 17
Factors that Affect
Solubility
Formation of Complex Ions
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Chapter 17
Factors that Affect
Solubility
Amphoterism
• Amphoteric oxides will dissolve in either a strong acid or
a strong base.
• Examples: hydroxides and oxides of Al3+, Cr3+, Zn2+, and
Sn2+.
• The hydroxides generally form complex ions with four
hydroxide ligands attached to the metal:
Al(OH3)(s) + OH-(aq)
Al(OH)4-(aq)
• Hydrated metal ions act as weak acids. Thus, the
amphoterism is interrupted:
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Chapter 17
Factors that Affect
Solubility
Amphoterism
• Hydrated metal ions act as weak acids. Thus, the
amphoterism is interrupted:
Al(H2O)63+(aq) + OH-(aq)
Al(H2O)5(OH)2+(aq) + H2O(l)
Al(H2O)5(OH)2+(aq) + OH-(aq)
Al(H2O)4(OH)2+(aq) + H2O(l)
Al(H2O)4(OH)2+(aq) + OH-(aq)
Al(H2O)3(OH)3(s) + H2O(l)
Al(H2O)3(OH)3(s) + OH-(aq)
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Al(H2O)2(OH)4-(aq) + H2O(l)
Chapter 17
Precipitation and
Separation of Ions
BaSO4(s)
Ba2+(aq) + SO42-(aq)
• At any instant in time, Q = [Ba2+][SO42-].
– If Q < Ksp, precipitation occurs until Q = Ksp.
– If Q = Ksp, equilibrium exists.
– If Q > Ksp, solid dissolves until Q = Ksp.
• Based on solubilities, ions can be selectively removed
from solutions.
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Chapter 17
Precipitation and
Separation of Ions
• Consider a mixture of Zn2+(aq) and Cu2+(aq). CuS (Ksp =
6  10-37) is less soluble than ZnS (Ksp =
2  10-25),
CuS will be removed from solution before ZnS.
• As H2S is added to the green solution, black CuS forms
in a colorless solution of Zn2+(aq).
• When more H2S is added, a second precipitate of white
ZnS forms.
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Chapter 17
Precipitation and
Separation of Ions
Selective Precipitation of Ions
• Ions can be separated from each other based on their salt
solubilities.
• Example: if HCl is added to a solution containing Ag+
and Cu2+, the silver precipitates (Ksp for AgCl is 1.8  1010) while the Cu2+ remains in solution.
• Removal of one metal ion from a solution is called
selective precipitation.
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Chapter 17
• Qualitative analysis is
designed to detect the
presence of metal ions.
• Quantitative analysis is
designed to determine
how much metal ion is
present.
Qualitative Analysis for
Metallic Elements
• We can separate a complicated mixture of ions into five
groups:
– Add 6 M HCl to precipitate insoluble chlorides (AgCl, Hg2Cl2,
and PbCl2).
– To the remaining mix of cations, add H2S in 0.2 M HCl to
remove acid insoluble sulfides (e.g. CuS, Bi2S3, CdS, PbS,
HgS, etc.).
– To the remaining mix, add (NH4)2S at pH 8 to remove base
insoluble sulfides and hydroxides (e.g. Al(OH)3, Fe(OH)3, ZnS,
NiS, CoS, etc.).
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Chapter 17
Qualitative Analysis for
Metallic Elements
– To the remaining mixture add (NH4)2HPO4 to remove insoluble
phosphates (Ba3(PO4)2, Ca3(PO4)2, MgNH4PO4).
– The final mixture contains alkali metal ions and NH4+.
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Chapter 17
End of Chapter 17
Additional Aspects of Aqueous
Equilibria
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Chapter 17